does anyone see any type of pattern in this? or is it totally random?
40
65
110
185
310
520
870
1450
2420
4040
6750
11270
18820
31430
52490
87660
146395
244480
408280
681825
$\displaystyle f(k) = 10^{\left(0.222716k + 1.60206 \right)} $Code:k f(k) error 0, 40.0, 40, 0.0 1, 66.8, 65, 1.8 2, 111.6, 110, 1.6 3, 186.3, 185, 1.3 4, 311.1, 310, 1.1 5, 519.6, 520, -0.4 6, 867.7, 870, -2.3 7, 1449.0, 1450, -1.0 8, 2419.9, 2420, -0.1 9, 4041.2, 4040, 1.2 10, 6748.8, 6750, -1.2 11, 11270.4, 11270, 0.4 12, 18821.6, 18820, 1.6 13, 31432.2, 31430, 2.2 14, 52491.7, 52490, 1.7 15, 87661.1, 87660, 1.1 16, 146394.1, 146395, -0.9 17, 244478.1, 244480, -1.9 18, 408278.4, 408280, -1.6 19, 681825.0, 681825, 0.0 20, 1138647.7 21, 1901541.7 22, 3175574.7 23, 5303209.7 24, 8856360.2 25, 14790121.5 26, 24699502.8
.
Not entirely random. If the function were to round to the nearest 5 value, then it looks as if your entire set could be matched.
From where did those numbers originate?
.
Here is the full table. These numbers are for a game. The 1-20 is the level of the building. So for each of these sets does it look like it is rounding to the nearest 5? But I dont understand, because take the first set for example.
1-2 Increase of 25
2-3 Increase of 45
3-4 Increase of 75
and so on.
I am trying to figure out if there is a formula or sequence they are following.
Sorry, not very good at math at all, so layman's terms would be helpful.
In essence what I want to do is copy their formula, but change the numbers slightly to fit our game. So I am wondering if there is some sort of sequence to it. Basically could I break those numbers down to give me a starting number. And then change that starting number, so the final number is different?
For example (and this is completely incorrect, but the best way I can describe it)
If to get to 40 they start with 5 , so 40/2 = 20, 20/2 = 10, 10/2 = 5
then to get to 65 they start with 65/2 = 32.5, 32.5/2 = 16.25, 16.25/2 = 8.125
ect. Is there a starting number they are using that we could then create a sequence of formula to come up with the increasing numbers.
Does this make any sense at all?
1 2 3 4
1 40 100 50 60
2 65 165 85 100
3 110 280 140 165
4 185 465 235 280
5 310 780 390 465
6 520 1300 650 780
7 870 2170 1085 1300
8 1450 3625 1810 2175
9 2420 6050 3025 3630
10 4040 10105 5050 6060
11 6750 16870 8435 10125
12 11270 28175 14090 16905
13 18820 47055 23525 28230
14 31430 78580 39290 47150
15 52490 131230 65615 78740
16 87660 219155 109575 131490
17 146395 365985 182995 219590
18 244480 611195 305600 366715
19 408280 1020695 510350 612420
20 681825 1704565 852280 1022740
Sum 1699410 4248545 2124275 2549120
NO. Some. Maybe. A little. Partially.Does this make any sense at all?
A few new questions:
What are columns 3, 4, & 5 supposed to represent?
You did not have the sequence numbers (levels 1-20) in your original post when you listed all of the numeric values in column 2.
Column 3 is greater than column 2; column 4 is less than column 3; and column 5 is greater than column 4.
Does the summation (at the bottom) have any particular value?
Do you want your numeric values to be mapped proportionately scaled?
Using the funtion f(k) posted above:
EXAMPLE 1
to get the number for your Level 12:
Since I started at zero, you started at 1.
Your LEVEL 12 is equal to my sequence number 11
$\displaystyle 11 \cdot 0.222716 = 2.449876 $
$\displaystyle 2.449876 \, + \, 1.60206 \, = \, 4.051936 $
$\displaystyle 10^{4.051936} = 11270.36 $
Rounding 11270.3 to the nearest 5 --> 11270
EXAMPLE 2
Your LEVEL 17 is equal to my sequence number 16
$\displaystyle 16 \cdot 0.222716 = 3.56346 $
$\displaystyle 3.56346 \, + \, 1.60206 \, = \, 5.16552 $
$\displaystyle 10^{5.16552} = 146393 $
Rounding 146393 to the nearest 5 --> 146395
~~~
It helps to understand the problem concept if all of the pertinent data is available initially.
Could you elaborate on the additional data you put in your last post?
.
lol. Well. Columns 1-4 are a specific type of resource cost. So they dont have to be proportionally scaled.
here is the spreadsheet in its entirety.
Each column represents a cost of "imaginary resources". As they are probably not related to each other, Im assuming the levels themselves must be somehow proportionately scaled. Does your caculation seem to work for the other columns as well? Again, I am terrible at math. But can use your calculations to come up with a excel formula.
What am I trying to do? Trying to figure out the formula of how they came up with the numbers in each column. Then once I figure out the formula, I can continue to divide the numbers to give me a start number. After I have a start number I can change that number by just a little bit to have the final number differ from what they have, but follow their same formula. This is really hard to explain, but i appreciate your patience and help trying to figure this out for me.
to give a little better idea.
say the numbers in column 1 were as follows:
5
10
20
40
80
etc
it would be very easy for me to break it down and just change it to
4
8
16
32
64
etc
but since i cant figure out if their formula is somehow sequenced or scaled, i cant create my own numbers. so hence why i am wondering if it it just randomly created.
i suppose I could also use the total sum and proportionately break it down through 20 levels. I myself and now confused. HAHA
This is a computed table:
,40,100,50,60,
,1.602059991,2,1.698970004,1.77815125,<-Log(of 1st Value)
,0.2227165,0.2227165,0.2227165,0.2227165,<-log increment
A, B, C, D, E
1,40,100,50,60
2,65,165,85,100
3,110,280,140,165
4,185,465,235,280
5,310,780,390,465
6,520,1300,650,780
7,870,2170,1085,1300
8,1450,3625,1810,2175
9,2420,6050,3025,3630
10,4040,10105,5050,6060
11,6750,16870,8435,10125
12,11270,28175,14090,16905
13,18820,47055,23525,28230
14,31430,78580,39290,47150
15,52490,131230,65615,78740
16,87660,219155,109575,131490
17,146395,365985,182995,219590
18,244480,611195,305600,366720
19,408280,1020695,510350,612420
20,681825,1704565,852280,1022740
a is the value in column A from 1 to 20
b is the value in column B matched with the value in column A.
$\displaystyle b = 10^{ 0.2227165 (a-1) + 1.602059991}$
$\displaystyle b = INT[ b/5 + 0.5] \cdot 5 $
'INT' is the integer function or floor function.
---
The following are the formulas for an excel spread sheet.
DecimalValue = 10^[(LEVEL - 1)*0.2227165 + log(FirstLevelNumber)]Code:ROW>A, B, C, D, E COL 01] 0, 40, 100, 50, 60 02] 0 03] 0, =LOG(B1), =LOG(C1), =LOG(D1), =LOG(E1), <-Log(of 1st Value) 04] 0 05] 0, 0.2227165, 0.2227165, 0.2227165, 0.2227165, <-log increment 06] 0 07] 1, =INT((10^(($A7-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A7-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A7-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A7-1)*E$5+E$3))/5+0.5)*5 08] 2, =INT((10^(($A8-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A8-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A8-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A8-1)*E$5+E$3))/5+0.5)*5 09] 3, =INT((10^(($A9-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A9-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A9-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A9-1)*E$5+E$3))/5+0.5)*5 10] 4, =INT((10^(($A10-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A10-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A10-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A10-1)*E$5+E$3))/5+0.5)*5 11] 5, =INT((10^(($A11-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A11-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A11-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A11-1)*E$5+E$3))/5+0.5)*5 12] 6, =INT((10^(($A12-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A12-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A12-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A12-1)*E$5+E$3))/5+0.5)*5 13] 7, =INT((10^(($A13-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A13-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A13-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A13-1)*E$5+E$3))/5+0.5)*5 14] 8, =INT((10^(($A14-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A14-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A14-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A14-1)*E$5+E$3))/5+0.5)*5 15] 9, =INT((10^(($A15-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A15-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A15-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A15-1)*E$5+E$3))/5+0.5)*5 16] 10, =INT((10^(($A16-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A16-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A16-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A16-1)*E$5+E$3))/5+0.5)*5 17] 11, =INT((10^(($A17-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A17-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A17-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A17-1)*E$5+E$3))/5+0.5)*5 18] 12, =INT((10^(($A18-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A18-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A18-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A18-1)*E$5+E$3))/5+0.5)*5 19] 13, =INT((10^(($A19-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A19-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A19-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A19-1)*E$5+E$3))/5+0.5)*5 20] 14, =INT((10^(($A20-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A20-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A20-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A20-1)*E$5+E$3))/5+0.5)*5 21] 15, =INT((10^(($A21-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A21-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A21-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A21-1)*E$5+E$3))/5+0.5)*5 22] 16, =INT((10^(($A22-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A22-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A22-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A22-1)*E$5+E$3))/5+0.5)*5 23] 17, =INT((10^(($A23-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A23-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A23-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A23-1)*E$5+E$3))/5+0.5)*5 24] 18, =INT((10^(($A24-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A24-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A24-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A24-1)*E$5+E$3))/5+0.5)*5 25] 19, =INT((10^(($A25-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A25-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A25-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A25-1)*E$5+E$3))/5+0.5)*5 26] 20, =INT((10^(($A26-1)*B$5+B$3))/5+0.5)*5, =INT((10^(($A26-1)*C$5+C$3))/5+0.5)*5, =INT((10^(($A26-1)*D$5+D$3))/5+0.5)*5, =INT((10^(($A26-1)*E$5+E$3))/5+0.5)*5
PointValue= INT(DecimalValue/5+0.5)*5
The log values are common logs, not the natural logs.
If you carefully enter the equation 1 time, you can replicate to all of the other cells.
Hope that helps.
If you have any questions or need additional information...
.
lol. that math is so far beyond anything i know. im lucky sometimes if i can remember multiplication tables. but if i look at things long enough ill figure them out. Ill have a question or two tomorrow. But the fact you already wrote the excel formulas? wow. thank you. Ill give them a try tomorrow.
amazed at how people like you understand math so well.
I'm very interested! I would love to find out more inforamtion related to this topic. Thanks in advance.
me too, I need more detailed info
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